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Considering the collisions only partially elastic won't help you solve your problem. CR of the balls is exactly the same when the incoming ball hits a single ball or a group of them. The problem is that you can forecast the exact outcome of a collision only when it takes place between two balls or in a fixed line of balls like in a Newton's cradle. When ...

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Here are two separate ways to address the issue you bring up. One is more mathematical---comparing the relations $mv$ and $\frac{1}{2}mv^2$. The other has more to do with force and energy, which I'm calling physical. Mathematical Let's imagine two objects that are moving in the same direction collide with each other. Just to keep things simple, let's also ...

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Let's take an example with simple numbers : 1+2=3 3+0=3 This can represent the momentum conservation. Now look at the sum of squares : 1*1+2*2=5 3*3+0*0=9 The sum is not conserved because the momentum that was transferred changed differently the result of the squares. In a word, kinetic energy doesn't change linearly with speed (which is obvious since ...

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We know that in an inelastic collision that total momentum of the system before collision equals the total momentum after collision. But total kinetic energy before collision is not equal to total kinetic energy after collision. How is possible given that the formula of momentum is $mv$ and the formula of kinetic energy is $\frac{1}{2} > ... 0 Physicist created momentum as the property of the system that is conserved if on the system act only internal forces. A force is defined internal in the system if, the force itself and its reaction are applied on the system. From the previous hypotesis you can demostrate that momentum is defined as $$\vec{p} = \sum m_i\vec{\upsilon}_i$$ The last example ... 0 An elastic collision means that the over all kinetic energy of the entire system before and after the collision is the same. So the ball can bounce off the wall, and the wall can recoil in such a manner that you have an elastic collision. 3 Momentum is conserved in magnitude and direction. So in order to analyze any situation of momentum conservation, you should always start with $$\sum \mathbf p_{i}=\sum\mathbf p_f$$ where the subscripts denote the initial and final momenta. As to the ball & wall, you are correct that momentum is not conserved if you are only looking at the ball. If you ... 2 It is the momentum of the entire system that is conserved. The fundamental reason for this is that the laws of physics are the same everywhere in space. This argument for momentum conservation is called Noether's Theorem. So where did you go wrong in your original example? Well you assumed that the wall was completely rigid. In reality that isn't actually ... 0 Okay, let's first review the initial setup: 1st particle: mass$m$and initial velocity$v_m>0$in +x direction 2nd particle: mass$Am$with assumption$A>0$, and initial velocity$v_A=0$3rd particle: mass$Bm$with assumption$B>0$, and initial velocity$v_B=0$Notation: after each collision, the new velocities will have a prime added to ... 0 Which scenario to chose? A) I am standing with my back against a massive granite wall. A solid block of concrete of mass 100,000 kg is approaching me with a momentum of 10,000 kg m/s. It follows that the block moves at a speed of 0.1 m/s with a kinetic energy of 500 J. I stretch out my arms and when the block reaches my hands, I push with a force of 500 N. ... -1 What causes damage (pain) is Energy (.../time, do you agree with Brandon?). Suppose the momentum of the bodies is 1000 kg*m/s if A has mass 1000 Kg its KE is 500 J If B has mass 1 kg its KE energy is 500 000 J Car B is more destructive (painful) than A -2 In a general sense, the amount of "pain" that someone feels is a result of the pressure exerted on them (like a bed of nails vs. a single nail). Momentum is defined as$p=mv$, and pressure is defined as$P=F/A$Because$F=ma$(Newton's 2nd law), and$a=\frac{\Delta v} {\Delta t}$, we can say that$P=\frac{m \Delta v} {A \Delta t}$. Assuming that you ... 2 Let us first calculate how much energy carried by an object of mass$m$and momentum$p$. So the velocity is$v=p/m$, and therefore, the kinetic energy is$T=mv^2/2=p^2/2m$Therefore, if$p$is a constant, the heavier the object is, the less the kinetic energy$T$it carries. This means that a heavier object hits you and you receive less energy, which ... 3 This is an answer to the question version 1. Later versions invalidate the details of this answer, but some of the ideas are still valid. Will edit to current version if I get a chance. I'll define "pain" as the change in momentum, or the energy delivered (the two are related by your velocity after the impact, provided your mass is unchanged, so unless you ... 5 Regardless of the physical undefinability of "painfulness", I'd like to plug some numbers in a particular scenario: Let's have a momentum of$p = 1000 $m$\cdot$kg/s, A 0.25kg bullet would be fatal, moving at$v = 1000/0.25 = 4000$m/s, while a 2000kg car moves at$v=0.5$m/s, So at least in this scenario and particularly for relatively low momentum ... -1 Let E = kinetic energy P = momentum Pr = pressure A = contact area E(a) = 1/2 m*v^2 => E(a) = 1/2 P * V(a) E(b) = 1/2 m*v^2 => E(b) = 1/2 P * V(b) if V(a) < V(b) => E(a) < E(b) Pr = F/A = (m*a) /A = (m * v/t)/A = (p/t)/A => Pr = P / ( A * t) If we assume that both vehicles have the same elasticity the time t (contact time) is the same ... 0 Eric Angle has it pretty much right. In an inelastic collision some of the kinetic energy is absorbed by the deformation of the material. For example, if two balls of putty collide and stick together, kinetic energy is absorbed by squishing the putty. In a second example, if you shoot a bullet at a log, some of the kinetic energy is absorbed by friction ... 0 You are correct. The total angular momentum remains constant. As the objects are not approaching head-on, there must be angular momentum. That momentum remains in the joined object as it spins around its COM. The speed of the spin for the joined object will be proportional to the perpendicular distance and speed of approach, and inversely proportional to ... 0 It's a mistake, it happens, the text says clearly :" ...if the collision is elastic...." (wiki:)The conservation of momentum is expressed by the equation$\,\! m_{1}\vec{u}_{1}+m_{2}\vec{u}_{2}=m_{1}\vec{v}_{1} + m_{2}\vec{v}_{2}$. If you consider m=1 the velocity of M is simply$v_M =\frac {\sqrt{8gl}}{M+1}\$

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This is a very interesting question. The problem is simpler in kinematics. However, if we view it as a problem in dynamics, invoking forces and Newton's laws, then the question becomes a natural consequence and the answers become rather complicated. The question is one of reference frames, in kinematics. The collision is viewed from the laboratory frame of ...

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In inelastic collisions, kinetic energy is not conserved, so I'm going to assume you mean a totally elastic collision since you say energy is conserved. O.K, so when the ball hits the wall, the speed of the wall before and after is 0, so that means the kinetic energy of the ball is conserved and thus the magnitude of the velocity is the same before and after ...

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